What is the extreme f(x,y)=16-x^2-y^2 ?

The extreme f(x,y)=16-x^2-y^2 is Maximum(0,0)

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extreme f(x,y)=16-x^2-y^2

extreme points

f (x, y) = 16 - x^{2} - y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 2

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x, y) = 3 y^{3} + 5 x^{2} y - 24 x^{2} - 24 y^{2} - 2

f (x, y) = 2 x^{3} + 3 y^{2} - 4 x y - 4

f (x) = 14 + 4 x - x^{2}, 0 \leq x \leq 5

- 1.8 x^{2} + 7 x - 11

f (x, y) = 2 ln (x) + ln (y) - 4 x - y

What is the extreme f(x,y)=16-x^2-y^2 ?
The extreme f(x,y)=16-x^2-y^2 is Maximum(0,0)